Question: The lifespans of bears in a particular zoo are normally distributed. The average bear lives $32.4$ years; the standard deviation is $4.2$ years. Use the empirical rule (68-95-99.7%) to estimate the probability of a bear living less than $40.8$ years.
$32.4$ $28.2$ $36.6$ $24$ $40.8$ $19.8$ $45$ $95\%$ $2.5\%$ $2.5\%$ We know the lifespans are normally distributed with an average lifespan of $32.4$ years. We know the standard deviation is $4.2$ years, so one standard deviation below the mean is $28.2$ years and one standard deviation above the mean is $36.6$ years. Two standard deviations below the mean is $24$ years and two standard deviations above the mean is $40.8$ years. Three standard deviations below the mean is $19.8$ years and three standard deviations above the mean is $45$ years. We are interested in the probability of a bear living less than $40.8$ years. The empirical rule (or the 68-95-99.7 rule) tells us that $95\%$ of the bears will have lifespans within 2 standard deviations of the average lifespan. The remaining $5\%$ of the bears will have lifespans that fall outside the shaded area. Because the normal distribution is symmetrical, half $({2.5\%})$ will live less than $24$ years and the other half $({2.5\%})$ will live longer than $40.8$ years. The probability of a particular bear living less than $40.8$ years is ${95\%} + {2.5\%}$, or $97.5\%$.